`x = 4-y^2 , x = y-2` Find the area of the region by integrating with (a) respect to x and (b) respect to y

`x=4-y^2 , x=y-2` 

(a) first let us find the area of the region with respect to x

so ,

the lines are

`x=4-y^2`

=> `y^2=4-x`

`y=sqrt(4-x)` -------------(1)

and

`x=y-2`

=> `y=x+2`---------(2)

let us find the curves where they intersect

so ,

`sqrt(4-x) =x+2`

`4-x=(x+2)^2`

`4-x=x^2+4x+4`

=>`x^2+5x=0`

=>` x(x+5)=0`

=>`x=0 or x=-5`

and the let let us get the points where they intersect with respect to y

`4-y^2=y-2`

`6=y+y^2`

=> `y^2+y-6=0`

=>`y^2+3y-2y-6=0`

=>`y(y+3)-2(y+3)=0`

=>`(y-2)(y+3)=0`

so `y=2 or y= -3`

so the points of intersection of the curves are `(0,2)` & `(-5,-3)`

but the curve `x=4-y^2` is beyond x=0 and intersects the x-axis (setting y=0) at x=4.

So the area of the region with respect to x -axis needs to have two integrals.

Area=`int_-5^0[(x+2)-(-(sqrt(4-x)))] dx + ` 

`int_0^4[(sqrt(4-x)-(-(sqrt(4-x)))]` `dx` 

=`int_-5^0 [(x+2)+((sqrt(4-x)))] dx`
`+int_0^4 [(sqrt(4-x)+((sqrt(4-x)))] dx`

=`[x^2/2+2x-2/3(4-x)^(3/2)]_-5^0 +[-2/3(4-x)^(3/2)-2/3(4-x)^(3/2)]_0^4 `

=`[(-2/3)(4)^(3/2)]-[25/2-10-(2/3)(9)^(3/2)]+[0]-[(-4/3)(4)^(3/2)]`

=`-16/3+31/2 +0+32/3`

=`61/6+32/3` =`125/6`

 so the Area is `125/6 `

`(b)` area of the region with respect to y is

    Area = `int _-3 ^2 [(y-2)-(4-y^2)] dy`

   = `[y^2/2 -2y -4y+y^3/3]_-3 ^2`

  =` [4/2 -4 -8 +8/3]-[9/2 +6+12-27/3]`

  `=-22/3 -27/2`

`=-125/6`

`= -20.833`

But since the area cannot be negative then the area is `20.833` or `125/6`